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प्रश्न
20 g of ice at 0°C absorbs 10,920 J of heat energy to melt and change to water at 50°C. Calculate the specific latent heat of fusion of ice. Specific heat capacity of water is 4200 J kg-1 K-1.
उत्तर
Let Specific Latent heat of fusion of ice = L
Heat required to change ice at 0° C to water 0°C + water 50°C
∴ Heat required = `20"L" + 20 xx 4200/1000 xx (50 - 0)`
= 10,920
Q = 20L + 4200
= 10,920
20L = 6720
L = `6720/20`
= 336 J g-1
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